If $A^4 = I$ (Identity Matrix) Then $A^2 = \pm I$? 
If $A^4 = I$ (Identity Matrix), is then $A^2 = \pm I$?

I believe the statement is true by doing a few examples, but I am struggling to start my proof. Any hints would be appreciated.
 A: You've done too few examples.
If you allow complex entries, consider $$ A=\begin{pmatrix}i&0\\0&1\end{pmatrix},$$
if not, consider
$$ A=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}.$$ 
A: Your question can be put more simply, by defining $B=A^2$, as: does $B^2=I$ imply $B=\pm I$?
And the answer is not. Consider first that $B^2=I \iff B=B^{-1}$.  Then, see here.
As pointed out in an aswer, the necessary and sufficient condition is to have $B$ diagonalizable and with eigenvalues $\pm 1$ [*] 
[*] Edit: As per David C. Ullrich's comment, strictly speaking we need an additional restriction: that $B$ has a square root, that is, there exists some $A$ such that $A^2=B$. That's automatically true for complex matrices, but not so for real matrices (or some other fields). For example, both $B=\begin{pmatrix}1 &0\\0 &-1\end{pmatrix}$ and $B=-I$  verify the $B^2=I$ condition, but 
they don't have square root in the reals.
Given that, in the question, you seem to accept the alternative $A^2=-I$, suggests that you are dealing with the complex field, but this needs to be pointed out.
